If you want a new array result containing a copy of arr whenever arr < 255, and 255 otherwise:

result = np.minimum(arr, 255)

More generally, for a lower and/or upper bound:

result = np.clip(arr, 0, 255)

If you just want to access the values over 255, or something more complicated, @mtitan8's answer is more general, but np.clip and np.minimum (or np.maximum) are nicer and much faster for your case:

In [292]: timeit np.minimum(a, 255)
100000 loops, best of 3: 19.6 µs per loop

In [293]: %%timeit
   .....: c = np.copy(a)
   .....: c[a>255] = 255
   .....: 
10000 loops, best of 3: 86.6 µs per loop

If you want to do it in-place (i.e., modify arr instead of creating result) you can use the out parameter of np.minimum:

np.minimum(arr, 255, out=arr)

or

np.clip(arr, 0, 255, arr)

(the out= name is optional since the arguments in the same order as the function's definition.)

For in-place modification, the boolean indexing speeds up a lot (without having to make and then modify the copy separately), but is still not as fast as minimum:

In [328]: %%timeit
   .....: a = np.random.randint(0, 300, (100,100))
   .....: np.minimum(a, 255, a)
   .....: 
100000 loops, best of 3: 303 µs per loop

In [329]: %%timeit
   .....: a = np.random.randint(0, 300, (100,100))
   .....: a[a>255] = 255
   .....: 
100000 loops, best of 3: 356 µs per loop

For comparison, if you wanted to restrict your values with a minimum as well as a maximum, without clip you would have to do this twice, with something like

np.minimum(a, 255, a)
np.maximum(a, 0, a)

or,

a[a>255] = 255
a[a<0] = 0

Assuming the values are between 0 and some maximum integer, one could implement a fast replace by using the numpy-array as int->int dict, like below

mp = numpy.arange(0,max(data)+1)
mp[replace.keys()] = replace.values()
data = mp[data]

where first

data = [ 0  1  6  5  1  2  7  6  2  3  8  7  3  4  9  8  5  6 11 10  6  7 12 11  7
  8 13 12  8  9 14 13 10 11 16 15 11 12 17 16 12 13 18 17 13 14 19 18 15 16
 21 20 16 17 22 21 17 18 23 22 18 19 24 23]

and replacing with

replace = {4: 0, 9: 5, 14: 10, 19: 15, 20: 0, 21: 1, 22: 2, 23: 3, 24: 0}

we obtain

data = [ 0  1  6  5  1  2  7  6  2  3  8  7  3  0  5  8  5  6 11 10  6  7 12 11  7
  8 13 12  8  5 10 13 10 11 16 15 11 12 17 16 12 13 18 17 13 10 15 18 15 16
  1  0 16 17  2  1 17 18  3  2 18 15  0  3]

Numpy provide the comparison to a scalar with the standard == operator, such that arr == v return a boolean array, often called a mask. The masked selection arr[arr == v] takes the subset of arr where the condition is met, so this snippet should work.

import numpy as np
arr = np.array([7, 1, 2, 0, 2, 3, 4, 0, 5])
arr[arr == 7] = 2
arr
array([2, 1, 2, 0, 2, 3, 4, 0, 5])

You would need to clip subsets of the arrays.

something like this:

a[2:].clip(0, None)

You could do this a couple of ways. One would be in a for loop:

for list in list_of_lists:
    list[2:] = list[2:].clip(0, None)

Or, using [:, 2:] which references your list of lists (:), and then all sublists of that (2:).

The result is basically what Joe Kingston has suggested:

list[:, 2:] = list[:, 2:].clip(0, None)

You have two errors. There are missing parenthesis in the first line of main:

x = np.array(([1,2,2], [1,4,3]))

And you must replace range(x) by range(len(x)) in the Iter_Replace function.

Instead of replacing the values one by one, it is possible to remap the entire array like this:

import numpy as np
a = np.array([1,2,2,1]).reshape(2,2)
# palette must be given in sorted order
palette = [1, 2]
# key gives the new values you wish palette to be mapped to.
key = np.array([0, 10])
index = np.digitize(a.ravel(), palette, right=True)
print(key[index].reshape(a.shape))

yields

[[ 0 10]
 [10  0]]

Credit for the above idea goes to @JoshAdel. It is significantly faster than my original answer:

import numpy as np
import random
palette = np.arange(8)
key = palette**2
a = np.array([random.choice(palette) for i in range(514*504)]).reshape(514,504)

def using_unique():
    palette, index = np.unique(a, return_inverse=True)
    return key[index].reshape(a.shape)

def using_digitize():
    index = np.digitize(a.ravel(), palette, right=True)
    return key[index].reshape(a.shape)

if __name__ == '__main__':
    assert np.allclose(using_unique(), using_digitize())

I benchmarked the two versions this way:

In [107]: %timeit using_unique()
10 loops, best of 3: 35.6 ms per loop
In [112]: %timeit using_digitize()
100 loops, best of 3: 5.14 ms per loop

you are missing ,s in the definition of your array. your array is the same as this: [["id11.0"], ["id20.0"]] (strings just get concatenated). if your arrays are numpy arrays then this is the way they are represented (printed). but that does not work as input...

starting from your code you could to this:

array = [["id1", "1.0"], ["id2", "0.0"]]

for i, (id_, number) in enumerate(array):
    if number == "0.0":
        array[i] = [id_, "ClassA"]
    else:
        array[i] = [id_, "ClassB"]

or, more elegant, use a list-comprehension:

array = [[id_, "ClassA"] if number == "0.0" else [id_, "ClassB"]  
         for id_, number in array ]